1.
[tex]dane:\\m = 0,5 \ kg\\\Delta T = 10^{o}C\\Q = 1,18 \ kJ = 1 \ 180 \ J\\szukane:\\c = ?\\\\c = \frac{Q}{m\cdot\Delta T}\\\\c = \frac{1180 \ J}{0,5 \ kg\cdot10^{o}C}\\\\\underline{c = 236\frac{J}{kg\cdot^{o}C}}[/tex]
Odp. To mogło być srebro.
2.
[tex]dane:\\m = 20 \ kg\\T_1 = 20^{o}C\\T_2 = 100^{o}C\\\Delta T = T_2-T_1=100^{o}C - 20^{o}C = 80^{o}C\\c = 4200\frac{J}{kg\cdot^{o}C}\\szukane:\\Q = ?\\\\Q = m\cdot c \cdot \Delta T\\\\Q = 20 \ kg\cdot4200\frac{J}{kg\cdot^{o}C}\cdot80^{o}C\\\\\underline{Q = 6 \ 720 \ 000 \ J = 6 \ 720 \ kJ = 6,72 \ MJ}[/tex]
[tex]1 \ kJ = 1000 \ J\\1 \ MJ = 1 \ 000 \ 000 \ J[/tex]
